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(A) According to Graham's Law of diffusion,the rate of diffusion $R$ of a gas is inversely proportional to the square root of its molar mass $M$: $\fr

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$O_2$

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(A) According to Graham's Law of diffusion,the rate of diffusion $R$ of a gas is inversely proportional to the square root of its molar mass $M$: $\frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}}$.Given: $R_1 = 28.7 \; cm^3 s^{-1}$ (for Hydrogen,$M_1 = 2.02 \; g/mol$),$R_2 = 7.2 \; cm^3 s^{-1}$.Rearranging the formula for $M_2$: $M_2 = M_1 \left( \frac{R_1}{R_2} ight)^2$.Substituting the values: $M_2 = 2.02 	imes \left( \frac{28.7}{7.2} ight)^2$.$M_2 \approx 2.02 	imes (3.986)^2 \approx 2.02 	imes 15.89 \approx 32.1 \; g/mol$.The molar mass of Oxygen $(O_2)$ is approximately $32.0 \; g/mol$. Therefore,the gas is Oxygen.

From a certain apparatus,the diffusion rate of hydrogen has an average value of $28.7 \; cm^3 s^{-1}$. The diffusion of another gas under the same conditions is measured to have an average rate of $7.2 \; cm^3 s^{-1}$. Identify the gas.

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